Limit of sum( (1 + 4i/n)^3(8/n), i = 1,...,n ) as n approaches infinity

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Limit of sum( (1 + 4i/n)^3(8/n), i = 1,...,n ) as n approaches infinity

[@dodgsonlluis]

It can be understood as a Riemann sum and thus calculated using an integral, the one of 8(1+4x)^3 between 0 and 1.

[@lllll-wz2pt]

BUT I think when 3 become more higher number, there may be some little error:)

[@CosmicJxsh]

You are simply the best

[@94fahim]

Thank you!

[@thecloudtrainer]

Awesome!!!